| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection verification |
| Difficulty | Moderate -0.5 This is a straightforward two-part question on 3D vector lines requiring standard techniques: (i) substituting a point into both line equations to verify intersection (routine verification), and (ii) using the scalar product formula to find the angle between direction vectors. Both parts are direct applications of standard methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -5+3\lambda = 1 \Rightarrow \lambda = 2\) | M1 | finding \(\lambda\) or \(\mu\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = 4-2 = 2\), so \((1,3,2)\) lies on 1st line | E1 | verifying two other coordinates for line 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(z = 2+0 = 2\), so \((1,3,2)\) lies on 2nd line | E1 | verifying two other coordinates for line 2 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Angle between \(\begin{pmatrix}3\\0\\-1\end{pmatrix}\) and \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\) | M1 | direction vectors only |
| \(\cos\theta = \frac{3\times2+0\times(-1)+(-1)\times0}{\sqrt{10}\sqrt{5}} = 0.8485\ldots\) | M1 A1 | allow M1 for any vectors |
| \(\Rightarrow \theta = 31.9°\) | A1 | or \(0.558\) radians |
| [4] |
## Question 6:
### Part (i)
$x = -5+3\lambda = 1 \Rightarrow \lambda = 2$ | M1 | finding $\lambda$ or $\mu$
$y = 3+2\times0 = 3$
$z = 4-2 = 2$, so $(1,3,2)$ lies on 1st line | E1 | verifying two other coordinates for line 1
$x = -1+2\mu = 1 \Rightarrow \mu = 1$
$y = 4-1 = 3$
$z = 2+0 = 2$, so $(1,3,2)$ lies on 2nd line | E1 | verifying two other coordinates for line 2
| [3] |
### Part (ii)
Angle between $\begin{pmatrix}3\\0\\-1\end{pmatrix}$ and $\begin{pmatrix}2\\-1\\0\end{pmatrix}$ | M1 | direction vectors only
$\cos\theta = \frac{3\times2+0\times(-1)+(-1)\times0}{\sqrt{10}\sqrt{5}} = 0.8485\ldots$ | M1 A1 | allow M1 for any vectors
$\Rightarrow \theta = 31.9°$ | A1 | or $0.558$ radians
| [4] |6 (i) Verify that the lines $\left. \mathbf { r } = \begin{array} { r } - 5 \\ 3 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { r } 3 \\ 0 \\ - 1 \end{array} \right)$ and $\left. \left. \mathbf { r } = \begin{array} { r } - 1 \\ 4 \\ 2 \end{array} \right) + \mu - \begin{array} { r } 2 \\ - 1 \\ 0 \end{array} \right)$ meet at the point ( $1,3,2$ ).\\
(ii) Find the acute angle between the lines.
\hfill \mbox{\textit{OCR MEI C4 Q6 [7]}}